Hi, Everybody, I had a quick and subjective question that may interest you based on the following data. "If the incoming freshman class size is about 179 each year, they only graduate approximately 95 each year and they only transfer 5 each year, Is the following attrition(drop out) data approximately fit?" 1st year 18.1%, 2nd year 3.9%, 3rd year 2.5% 4th year 0 % and so on
When I found the directrix of the parabola with the equation -8(y-3)=(x+4)^2, I got y=1, but the correct answer is actually y=5.
What I did was add 1/4a to the y-coordinate of the vertex of the parabola to the parabola, since the parabola is pointing downwards, thus the directrix should be above it.
If anyone can tell me what I did wrong that would be great
@DarkRunner: i had to work this out (cuz I never remember it), but the directrix for $y=ax^2$ should be the line $y=-\frac 1{4a}$. So if your $a = -1/8$, it would be a distance $2$ above the vertex. That puts it at $y=5$.
My Doubt
Here $||f||_{1}=\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|$ where
as $||f||_0=\sup_{x\in[0,1]}|f(x)|$. We can easily prove from
definition that
$$||f||_0=\sup_{x\in[0,1]}|f(x)|\leq
\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|=||f||_1$$
Using the above inequality, ...
To check it is a weak extremum, you consider all $y$ with $\|y-\hat y\|_1$ small. This implies that $\|y-\hat y\|_0$ is small, and so it meets the definition of a strong extremum. Therefore, any strong extremum must be a weak extremum.
To check it is a weak extremum, you consider all $y$ with $\|y-\hat y\|_1$ small. This implies that $\|y-\hat y\|_0$ is small I am clear till here.
Suppose $\hat y$ is strong extremum then $\exists \epsilon>0$ . $||y-\hat y||_0<\epsilon$. how can we guarentee that there exists a positive number $\delta>0$: $||y-\hat y||_1<\delta$ from this?
If $\hat y$ is a strong extremum, then for every $y$ with $\|y-\hat y\|_0<\varepsilon$ (fixed), something happens. Then for that same $\varepsilon$, the something will still happen for every $y$ with $\|y-\hat y\|_1<\varepsilon$.
To answer your question directly, you take $\delta=\varepsilon$.
@TedShifrin the way I interpret the problem is that they are asking you find the distribution function for $X$ and $Y$ for the special case where $x^2 + y^2 < a^2$
A lot of Schaum's outlines were written 100 years ago and are not necessarily the best reference. You might try looking at a nice book by Jim Putman, who actually is a statistician. He has a lot of good applications in his book.
Yes, check your substitution in the integral. The $K$ should have canceled.
That's why it's there in the first place ... to have a probability distribution.
The probability is being in the entire plane must be $1$. That's why that $K$ is there in the first place. Probability density functions must always have total integral $1$.
We have the problem \begin{align*}&u_t+uu_x=0 \\ &u(x,0)=\begin{cases}2, & x<0 \\ 2-x, & x\in[0,1] \\1, & x>1\end{cases}\end{align*} We have the problems $$\frac{dx}{dt}=u \ \text{ and } \ \frac{du}{dt}=0$$ From the first one we get $x=ct+c_2$ and for $t=0$ we get $x_0=c_2$ and so we get $x_0=x-...
@TedShifrin 1)Suppose $\hat y$ is strong extremum then $\exists \epsilon>0$ . $||y-\hat y||_0<\epsilon$. 2)Suppose $\hat y$ is weak extremum then $\exists \delta>0$ . $||y-\hat y||_0<\delta$. $\delta=\epsilon$ I am clear. What happen $\delta>\epsilon$ and $\delta<\epsilon$?
If $\delta=\epsilon$ , $\hat y$ a strong extrema $\implies$ it is weak extrema.
@TedShifrin Thank you. Now it is clear, Sorry, I understood in a different way. My english is weak.
My Doubt
Here $||f||_{1}=\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|$ where
as $||f||_0=\sup_{x\in[0,1]}|f(x)|$. We can easily prove from
definition that
$$||f||_0=\sup_{x\in[0,1]}|f(x)|\leq
\sup_{x\in[0,1]}|f(x)|+\sup_{x\in[0,1]}|f'(x)|=||f||_1$$
Using the above inequality, ...
